The Unapologetic Mathematician

Mathematics for the interested outsider

Differentiation Under the Integral Sign

Another question about “partial integrals” is when we can interchange integral and differential operations. Just like we found for continuity of partial integrals, this involves interchanging two limit processes.

Specifically, let’s again assume that f:R\rightarrow\mathbb{R} is a function on the rectangle R=[a,b]\times[c,d], and that \alpha is of bounded variation on [a,b] and that the integral

\displaystyle F(y)=\int\limits_a^bf(x,y)\,d\alpha(x)

exists for every y\in[c,d]. Further, assume that the partial derivative \left[D_2f\right](x,y) exists and is continuous throughout R. Then the derivative of F exists and is given by

\displaystyle\frac{d}{dy}F(y)=\frac{d}{dy}\int\limits_a^bf(x,y)\,d\alpha(x)=\int\limits_a^b\frac{\partial}{\partial y}f(x,y)\,d\alpha(x)

Similar results hold where x and y are vector values, and where derivatives in terms of y are replaced outside the integral by partial derivatives in terms of its components.

So, if y_0\in(c,d) and y\neq y_0 we can calculate the difference quotient:


where \xi is some number between y_0 and y that exists by the differential mean value theorem. Now to find the derivative, we take the limit of the difference quotient

\displaystyle\begin{aligned}\frac{d}{dy}F(y_0)&=\lim\limits_{y\rightarrow y_0}\frac{F(y)-F(y_0)}{y-y_0}\\&=\lim\limits_{y\rightarrow y_0}\int\limits_a^b\left[D_2f\right](x,\xi)\,d\alpha(x)\\&=\int\limits_a^b\lim\limits_{y\rightarrow y_0}\left[D_2f\right](x,\xi)\,d\alpha(x)\end{aligned}

as we take y to approach y_0, the number \xi gets squeezed towards y_0 as well. Since we assumed \left[D_2f\right](x,y) to be continuous, the limit in the integrand will equal \left[D_2f\right](x,y_0), as asserted.

January 13, 2010 Posted by | Analysis, Calculus | 15 Comments